I want to solve the following integration
$$I = \int_0^\pi\int_0^{2\pi}\exp{\bigg[x\cos(\phi)\sin(\theta)+y\sin(\phi)\sin(\theta)+z\cos(\theta))\bigg]}\sin(\theta)\,d\phi \,d\theta $$My Attempt:

First solve the $\phi$ part
$$I =
\int_0^\pi\exp{[z\cos(\theta)}]\sin(\theta) \Bigg[\int_0^{2\pi}
\exp\bigg[x\cos(\phi)\sin(\theta)+y\sin(\phi)\sin(\theta))\bigg] \, d\phi\Bigg] \, d\theta\\
I = \int_0^\pi\exp[z\cos(\theta)]\sin(\theta) \, d\theta I_2
$$
where
$$I_2 = \int_0^{2\pi}
\exp\bigg[x\cos(\phi)\sin(\theta)+y\sin(\phi)\sin(\theta))\bigg] \, d\phi$$
Nothing seems to work here. I have tried integration by parts and substitution method but both just keeps expanding the terms. How can I solve this. Please help.

1 Answer
1

You may notice that the integral is in fact
$$
\iint_{S} e^{(\vec{v} \cdot \vec{r})} d\vec{r}, \quad \vec{v} = \{x;y;z\},
$$
where integration is performed over a sphere of unit radius.
You can use the symmetry of the problem and rotate everything so that $\vec{v}$ points along $Z$ axis.
This simplifies the expression drastically
$$
\int_0^{2\pi} \int_0^{\pi} e^{v\, cos(\theta)} \sin(\theta) d\theta d\phi, \quad
v = \sqrt{x^2 + y^2 + z^2}.
$$
The net result should be something like
$$
\frac{4 \pi \sinh(\sqrt{x^2+y^2+z^2})}{\sqrt{x^2+y^2+z^2}}.
$$

$\begingroup$@guest Thank you very much. Sir, how can someone be as genius as you? I was struggling with this question for 3 days and you have solved it in so simple way$\endgroup$
– Sana UllahMar 8 at 14:13